mechanics 2 : centres of mass

brief power point with worked examples of centre of mass questions and info on how to find CoM. includes: -discrete groups of particles in 1D, discrete groups of particles in 2D, laminas,composite shapes, and laminas in equilibrium. 

for OCR MEI M2 :) hope it helps 

HideShow resource information
  • Created by: M
  • Created on: 24-10-12 16:09
Preview of mechanics 2 : centres of mass

Other slides in this set

Slide 2

Preview of page 2

Here's a taster:

Discrete groups of particles in 1D
Three particles are placed at positions along the x-axis as shown.
Find the distance of the centre of mass of the group of particles
from O.
2m 3m 2m
m1= 3kg O m2= 1.5kg m3= 0.5kg
Use the formula mx=xM
m1 x+ m2 x+ m3 x=x(m1 + m2 + m3)
(3X-2)+(1.5X3)+(0.5X5)=x (3+1.5+0.5)
1=5x
x=0.2 therefore the centre of mass is 0.2m from O
By MI…read more

Slide 3

Preview of page 3

Here's a taster:

Discrete groups of particles in 2D
Find the centre of mass of the system of particles shown in the
diagram.
Use the formula: m(x,y)=(x,)M
2
1 m1 (x,y)+m2(x,y)+m3(x,)=(x,)(m1 +m2+m3)
-2 -1 1 2 1(-2,0)+6(-1,2)+3(1,1)=(x,)(1+6+3)
-1
(-5,15)=10(x,)
m1=1kg m2=6kg m3=3kg
(x,)=(-0.5,1.5)
If you are asked to find out the mass or position of one of the masses
just put in the values you know into the equation and rearrange to find
the unknown.
By MI…read more

Slide 4

Preview of page 4

Here's a taster:

Standard Uniform Laminas
Uniform laminas have evenly spread mass, so the centre of mass
is in the centre of the shape, on all the lines of symmetry. For
shapes with more than one line of symmetry, the centre of mass
is where the lines of symmetry intersect.
To find the centre of mass of a uniform triangle draw on the
medians (line from each vertex to midpoints of opposite side)
and the CoM is where they cross (the centroid.)
By MI…read more

Slide 5

Preview of page 5

Here's a taster:

Composite shapes
A composite shape is one that can be broken up into standard
parts such as rectangles and circles. Once you've found the CoM
of apart, imagine replacing It with a particle of the same mass in
the position of the CoM. Do this for each part, then find the CoM
of the group of `particles'- this is the CoM of the composite
shape.
You may have a shape that looks like a `standard' shape with
other standard shapes `removed' rather than stuck together. The
removal method is like the one above, except the individual
centres of mass are subtracted rather than added.
CGP, A2-level ,Mathematics ,Exam Board: OCR MEI, complete revision and practice page 144…read more

Slide 6

Preview of page 6

Here's a taster:

Laminas in equilibrium
Laminas hang with the centre of mass directly below the pivot.
In this shape, A is at (0.6),C is at (8,6) and CoM is at (4,5). Find in radians
to 3s.f., the angle AC makes with the vertical when the shape is
suspended from A.
B Draw everything you are told in the
question including a line representing
the vertical from the suspension point
4
through the CoM
A
C
1
Now just use trig to find the angle:
= tan-1 1/4 =0.245 radians to 3 s.f.
D
By MI…read more

Slide 7

Preview of page 7
Preview of page 7

Comments

No comments have yet been made

Similar Mathematics resources:

See all Mathematics resources »See all resources »